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Mathematics 22 Online
OpenStudy (lena772):

Help quick pleaseeeee!

OpenStudy (lena772):

ganeshie8 (ganeshie8):

start by finding y'

OpenStudy (lena772):

y'=6x^2+8x-5

ganeshie8 (ganeshie8):

Yes, find the slope at x = 1

OpenStudy (anonymous):

@ganeshie8 after you help this person with her question can you please help me with mine

ganeshie8 (ganeshie8):

sorry i suck at english :|

OpenStudy (lena772):

8?

ganeshie8 (ganeshie8):

y'(x) =6x^2+8x-5 slope at x = 1 y'(1) = 6(1)^2 + 8(1) - 5 = 6 + 8 -5 = 9

OpenStudy (lena772):

so what is the y intercept then

ganeshie8 (ganeshie8):

so slope, m = 9 u knw a point (1, -2) u can write teh equation of tangent in point slope form

OpenStudy (lena772):

y-1=9(x--2) y-1=9(x+2)

ganeshie8 (ganeshie8):

nope, do it in ur regular way may be

ganeshie8 (ganeshie8):

y = mx + b y = 9x + b

ganeshie8 (ganeshie8):

solve b, by putting (1, -2) above

OpenStudy (lena772):

-2=9*1+b -2=9+b -11=b

ganeshie8 (ganeshie8):

Yes, so the equation of tangent line at (1, -2) is :- y = 9x - 11

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